What is the tangent of y=(4x)/((x^2+1)) ?

The tangent of y=(4x)/((x^2+1)) is y=(4(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(8a_{0}^3)/((a_{0)^2+1)^2}

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tangent of y=(4x)/((x^2+1))

tangent of

y = \frac{4 x}{( x ^{2} + 1 )}

y = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{4 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{4 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{4 x}{x ^{2} + 1}

x = a_{0} : m = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

\frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{4 a _{0}}{a _{0}^{2} + 1}) : y = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

\int \frac{2}{9 x ^{2} - 1} d x

\frac{d}{d x} (3 + sin (2 x))

x^{2} - 4 x, 2 x

\int 1 + 2 x d x

9 y

What is the tangent of y=(4x)/((x^2+1)) ?
The tangent of y=(4x)/((x^2+1)) is y=(4(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(8a_{0}^3)/((a_{0)^2+1)^2}